Bug #9776
closed
Ruby double-splat operator unexpectedly modifies hash
Description
I noticed what I find to be a very surprising behavior with the double-splat (**) operator in Ruby 2.1.1.
When key-value pairs are used before a **hash, the hash remains unmodified. However, when key-value pairs are only used after the **hash, the hash is permanently modified.
h = { b: 2 }
{ a: 1, **h } # => { a: 1, b: 2 }
h # => { b: 2 }
{ a: 1, **h, c: 3 } # => { a: 1, b: 2, c: 3 }
h # => { b: 2 }
{ **h, c: 3 } # => { b: 2, c: 3 }
h # => { b: 2, c: 3 }
For comparison, consider the behavior of the splat (*) operator on arrays:
a = [2]
[1, *a] # => [1, 2]
a # => [2]
[1, *a, 3] # => [1, 2, 3]
a # => [2]
[*a, 3] # => [2, 3]
a # => [2]
The array remains unchanged throughout.
Tsuyoshi Sawada has also highlighted that the expression's result is the self-same object as the original hash:
h.object_id == { **h, c: 3 }.object_id # => true
I investigated parse.y to try to determine the error there, but I couldn't narrow it down any further than the list_concat or rb_ary_push function calls in the assocs : block of the grammar.
Without exhaustively examining the C source, I think the best clue to the mechanism behind the erroneous behavior might be the following:
h = { a: 1 }
{ **h, a: 99, **h } # => {:a=>99}
That we don't see {:a=>1} illustrates that h[:a] is already overwritten by the time the second **h is evaluated.
Here is the use case that led me to this discovery:
def foo (arg) arg end
h = { a: 1 }
foo(**h, b: 2)
h # => { a: 1, b: 2 }
In the above example, I don't want { b: 2 } permanently added to my existing hash. I'm currently solving it like this:
h = { a: 1 }
foo(**h.dup, b: 2)
h # => { a: 1 }
The call to #dup feels unnecessary, and is inconsistent with the analogous behavior when using the single * operator. If this bug is fixed, I'll be able to eliminate that call.
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